Difference between revisions of "1988 AIME Problems/Problem 6"
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a + 2b &=& 113</cmath> | a + 2b &=& 113</cmath> | ||
− | gives <math>a = 13</math>, <math>b = 50</math>. Now it is simple to calculate <math>(4,3)</math>. One way to do it is to see that <math>(2,2)</math> has <math>206 - | + | gives <math>a = 13</math>, <math>b = 50</math>. Now it is simple to calculate <math>(4,3)</math>. One way to do it is to see that <math>(2,2)</math> has <math>206 - 2b = 106</math> and <math>(4,2)</math> has <math>186</math>, so <math>(3,2)</math> has <math>\frac{106 + 186}{2} = 146</math>. Now, <math>(3,0)</math> has <math>3b = 150</math>, so <math>(3,2) = \frac{(3,0) + (3,4)}{2} = \Longrightarrow (3,4) = * = 142</math>. |
=== Solution 2 (general) === | === Solution 2 (general) === |
Revision as of 01:32, 6 June 2014
Problem
It is possible to place positive integers into the vacant twenty-one squares of the square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*).
Solution
Solution 1 (specific)
Let the coordinates of the square at the bottom left be , the square to the right , etc.
Label the leftmost column (from bottom to top) and the bottom-most row (from left to right) . Our method will be to use the given numbers to set up equations to solve for and , and then calculate .
We can compute the squares at the intersections of two existing numbers in terms of and ; two such equations will give us the values of and . On the fourth row from the bottom, the common difference is , so the square at has a value of . On the third column from the left, the common difference is , so that square also has a value of . Equating, we get .
Now we compute the square . By rows, this value is simply the average of and , so it is equal to . By columns, the common difference is , so our value is . Equating, .
Solving
\begin{eqnarray*}4b - 3a &=& 161\\ a + 2b &=& 113 (Error compiling LaTeX. ! LaTeX Error: \begin{eqnarray*} on input line 20 ended by \end{document}.)
gives , . Now it is simple to calculate . One way to do it is to see that has and has , so has . Now, has , so .
Solution 2 (general)
First, let the number to be placed in the first column, fourth row. Let the number to be placed in the second column, fifth row. We can determine the entire first column and fifth row in terms of and :
Next, let the number to be placed in the second column, fourth row. We can determine the entire second column and fourth row in terms of , , and :
We have now determined at least two values in each row and column. We can finish the table without introducing any more variables:
We now have a system of equations.
Solving, we find that . The number in the square marked by the asterisk is
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.